Vragenuurtje
Organische Chemie 1
ORC-12803
© 2006 Thomson Higher Education
Question I
Hybridisation of Carbon Atoms
sp3
sp2
sp
C-atom with 4 substituents
(only single bonds)
C-atom with 3 substituents
(double bond)
C-atom with 2 substituents
(triple bond)
109o
T1
120o
180o
Carbocation Structure and Stability
Why Markovnikov’s Rule works
Structure of carbocations
•
Carbocations are planar
•
The trivalent carbon is sp2
hybridized and has a
vacant p orbital
perpendicular to the plane
of the carbon
•
Carbon has three attached
groups
T5
Drawing Resonance Forms
• In general any three-atom grouping with
a p orbital on each atom has two
resonance forms:
• The actual structure of de resonance hybrid is
closer to the more stable form
T2
Question I
2
3
7
1
2
Question I
Noncovalent Interactions between Molecules
Noncovalent interactions
Also called intermolecular forces or van der Waals forces
•
•
•
Dipole-dipole forces (between polar molecules)
Dispersion forces (between all neighboring molecules)
Hydrogen bonds
A weak attraction between a hydrogen atom bonded to an
electronegative O or N and an electron lone pair on another
O or N atom
Strong dipole-dipole interaction involving polarized O−H
and N−H bonds
acceptor
acceptor
donor
T2
donor
Question I
YES
The hydrogen bonded to the electronegative O atom
can form a hydrogen bond with an unshared electron
pair on another O or N atom.
NO
There is no hydrogen bonded to the electronegative O
atoms.
NO
There is no hydrogen bonded to the electronegative O
atoms.
Question I
Polar Covalent Bonds: Electronegativity
• Polar covalent bonds
• A covalent bond in which the electron distribution between
atoms is unsymmetrical
•
T2
Bond polarity due to difference in electronegativity (EN)
Polar Covalent Bonds: Electronegativity
Electronegativity (EN)
• The ability of an atom to attract shared electrons in a
covalent bond
• Generally increases across the periodic table from
left to right and from bottom to top
T2
Polar Covalent Bonds: Dipole Moment
•
Lone-pair electrons on
oxygen and nitrogen stick
out into space away from
positively charged nuclei
giving rise to a
considerable charge
separation and
contributing to the dipole
moment
• Symmetrical structures of
molecules cause the
individual bond polarities
and lone-pair
contributions to exactly
cancel
T2
Question I
NO
The dipole moment is directed towards the
electronegative Cl atom.
YES
The individual bond polarities and lone pair
contributions exactly cancel.
Electronegativity of relevant elements:
H:
C:
Cl:
O:
2.1
2.5
3.0
3.5
Question II
Rules for Resonance Forms
Rule 1 – Individual resonance forms are imaginary, not real
• Real structure is a composite
Rule 2 – Resonance forms differ only in the placement of
their p or nonbonding electrons
T2
Rules for Resonance Forms
Rule 3 – Different resonance forms of a substrate do not
have to be equivalent
T2
Rules for Resonance Forms
Rule 4 – Resonance forms obey normal rules of valency
(follow the octet rule)
Rule 5 – The resonance hybrid is more stable than any
individual resonance form
• Resonance leads to stability
T2
Question II
Question II
I
IV
II
III
First to consider: the more stable the anion, the less likely it is to accept
a H+, so the most stable anion is the weakest base.
Resonance?
Anions II and III are resonance-stabilised
Electronegativity of charge-bearing atom
O:
3.5
N:
3.0
C:
2.5
So, between II and III, III can accept a negative charge better
So, between I and IV, IV can accept a negative charge better
Question III
Conformations of Other Alkanes
T3
Staggered conformation is found on a local energy minimum
Eclipsed conformation is found on a local energy maximum
Question III
D
A
B
C
Question III
Axial & Equatorial Bonds in Cyclohexane
Chair conformation of cyclohexane
• 2 positions for substituents on ring
• 6 Axial positions – vertical C−H bond
•
6 Equatorial positions – orientation ~30º from plane
axial positions
equatorial positions
NB: By ring flip axial & equatorial positions rapidly interchange!
T3
Question III
Most stable
Question IV
Orientation of Electrophilic Addition:
Markovnikov’s Rule
Regiospecific reactions - reactions where only one of
two possible orientations occurs
Markovnikov’s Rule (Vladimir Markovnivov,1869)
• In the addition of HX to an alkene, the H attaches to the
carbon with fewer alkyl substituents and the X attaches to
the carbon with more alkyl substituents
T5
Question IV
Question IV
Halogenation of Alkenes
Halogenation of cycloalkenes
• Only trans-stereoisomer of dihalide product is formed
• Reaction occurs with anti stereochemistry – the two
halogen atoms come from opposite faces of doublebond, one from top face and one form bottom face
T6
Halogenation of Alkenes
Reaction occurs through an intermediate bromonium
ion (R2Br+), formed by interaction of the alkene with
Br2 and simultaneous loss of Br−
T6
Question IV
Chloronium ion
intermediate
Question IV
anti
Oxidation of Alcohols
How to recognize oxidation/reduction reactions in
organic chemistry!
Oxidation state
•
•
a measure of the degree of oxidation of an atom in a chemical compound
the hypothetical charge that an atom would have if all bonds
to atoms of different elements were 100% ionic
Oxidation
•
•
the increase in oxidation state of an atom or atoms in the product
compared with the starting compound
a net loss of electrons to the oxidator
(Reduction: a net gain of electrons)
T8
Oxidation of Alcohols
How to calculate oxidation states in a Lewis structure!
Oxidation state
The difference between the number of valence electrons that a neutral atom
of that element would have and the number of electrons that "belong" to it in
the Lewis structure using a set of rules.
• electrons in a bond between atoms of different elements belong to the
most electronegative atom
• electrons in a bond between atoms of the same element are split equally
• electrons in a lone pair belong only to the atom with the lone pair
loss 2 e−
T8
loss 2 e−
Question IV
Note that the oxidation state of C1 and C2 are identical
(1+1) + 1 + 2 = 5 electrons
1 + 1 + 2 = 4 electrons
Ox. State = 4 − 5 = −1
Ox. State = 4 − 4 = 0
Oxidation state from C1 and C2 increases both from −1 to 0,
so oxidation reaction
Question V
The Reason for Handedness in Molecules: Chirality
Chirality center
• A carbon atom bonded to four different groups in an
organic molecule
• Most common cause of chirality
•
T4
Chirality is a property of the entire molecule
Sequence Rules for Specifying Configuration:
R,S convention
Configuration
The three-dimensional arrangement of substituents at a chirality center
1) Make priorities for 4 substituents (similar to E/Z)
2) Place lowest priority at the back
3 remaining substituents:
• R configuration
(1 → 2 → 3) clockwise
• S configuration
(1 → 2 → 3) counterclockwise
T4
Sequence Rules for Specifying Configuration
Configuration
•
The three-dimensional arrangement of substituents at
a chirality center
Sequence rules to specify the configuration of a chirality center:
Look at the four atoms directly attached to the chirality
center and assign priorities in order of decreasing atomic
number
1.
•
T4
The atom with the highest atomic number is ranked first; the atom
with the lowest atomic number (usually hydrogen) is ranked fourth
Sequence Rules for Specifying Configuration
2.
T4
If a decision cannot be reached by ranking the first atoms
in the substituents, look at the second, third, or fourth
atoms outward until a difference is found
Sequence Rules for Specifying Configuration
3.
T4
Multiple-bonded atoms are equivalent to the same number
of single-bonded atoms
Question V
‘S’ but H is pointing forward, so R
S
4
2
1
1
2
3
3
4
22 = 4
No, meso compounds are compounds that contain chirality
centers, but possess a plane of symmetry. Because the
substituents on the chiral carbons are different in
streptimidone it is impossible for any stereoisomer to have
a plane of symmetry.
Diastereomers
A molecule with n chirality centers can have up to 2n
stereoisomers (although it may have fewer)
Diastereomers
• They are stereoisomers that are not mirror images
2R, 3R stereoisomer and 2R, 3S stereoisomer
• They are stereoisomers but not enantiomers
•
Enantiomers have opposite configurations at all chirality
centers
Diastereomers have opposite configurations at one or more of
the chirality centers but not at all
T4
Meso Compounds
Tartaric acid: 2 chirality centers
Meso form
identical  not enantiomers!
the molecule has a plane of symmetry
Tartaric acid exists as only 3 stereoisomers:
two enantiomers & one meso form
T4
Question V
Sequence Rules: The E,Z Designation
E,Z system
Sequence rules used to assign priorities to the substituent
groups on the double-bond carbons (alkenes)
Example
1
1
Z configuration
2
2
1
1
2
2
Z = for German zusammen “together”
E = for German entgegen “opposite”
Z
T5
Question V
2
2
2
1
1
1
Z
1
2
E
2
1
2
1
E
Question VI
Reactions of Alkyl Halides:
Nucleophile/Lewis base with an alkyl halide
nucleophile
base
T8
The SN2 Reaction
The mechanism of the SN2 reaction
• The reaction takes place in a single
step
• Incoming nucleophile approaches
from a direction 180º away from the
leaving halide ion, thereby inverting
the stereochemistry at carbon
T7
Characteristics of the SN2 Reaction
The Substrate:
Steric Effects
A bulky substrate prevents the easy approach of the
nucleophile, making bond formation difficult
T7
The SN1 Reaction
Mechanism of the SN1:
• Involves more than
one step
• The first step
(spontaneous,
unimolecular
dissociation of the
alkyl bromide to yield
a carbocation) is ratelimiting
T7
Characteristics of the SN1 Reaction
The Substrate
The more stable the carbocation intermediate, the
faster the SN1 reaction
Secondary allylic and secondary benzylic carbocations
are about as stable as a tertiary alkyl carbocations
(due to resonance stabilization)
T7
E2 Reaction
E2 Reaction
• Occurs when an alkyl halide is
treated with a strong base
• Reaction takes place in a single
step through a transition state in
which the double bond begins to
form at the same time the H and
X groups are leaving
T8
E1 Reaction
E1 Reaction
•
•
•
T8
Two steps
First step rate-limiting
Carbocation
intermediate is present
Question VI
tertiary
Bromide gets substituted by a H3CNH group
Alkyl halide is a tertiary alkyl halide
- sterically blocked for a SN2
- will give a stable tertiary carbocation intermediate in a SN1
Question VI
Question VI
Question VI
Question VI
X
secondary
secondary
methyl
secondary
SN2 reaction occurs in a single step:
- Nucleophile approaches the C−X bond from the back
- Bulky substituents will block the nucleophile’s approach
Question VII
Aromaticity and the Hückel 4n + 2 Rule
The Hückel 4n + 2 rule
A molecule is aromatic only if it has a planar, cyclic conjugated p
electron system which contains a total of 4n + 2 p electrons, where n is
an integer (n = 0, 1, 2, 3,…). The p electrons are completely delocalized
around the ring (resonance forms, more stability)
n = 0 (2 el.)
T9
n = 1 (6 el.)
Aromatic Heterocycles
Pyridine
The nitrogen atom is also sp2-hybridized and has one
electron in a p orbital, bringing the total to six p electrons
The nitrogen lone pair electrons are in an sp2 orbital in the plane of
the ring and are not involved with the aromatic p system
T9
Aromatic Heterocycles
Pyrrole is a five membered heterocycle with six p electrons
•
•
•
T9
Aromatic
Each of the sp2-hybridized carbons contributes one p electron
The sp2-hybridized nitrogen atom contributes the two electrons
from its lone pair, which occupies a p orbital
Question VII
X
X
Alkylation and Acylation of Aromatic Rings
Mechanism of the Friedel-Crafts alkylation reaction
•
The electrophile is
a carbocation,
generated by
AlCl3-assisted
dissociation of an
alkyl halide
•
Take care of
carbocation
rearrangement,
particularly when a
primary alkyl
halide is used
T9
Question VII